This question is about implementing a basic elimination algorithm for Candy Crush.
Given an m x n integer array board representing the grid of candy where board[i][j] represents the type of candy. A value of board[i][j] == 0 represents that the cell is empty.
The given board represents the state of the game following the player's move. Now, you need to restore the board to a stable state by crushing candies according to the following rules:
- If three or more candies of the same type are adjacent vertically or horizontally, crush them all at the same time - these positions become empty.
- After crushing all candies simultaneously, if an empty space on the board has candies on top of itself, then these candies will drop until they hit a candy or bottom at the same time. No new candies will drop outside the top boundary.
- After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
- If there does not exist more candies that can be crushed (i.e., the board is stable), then return the current board.
You need to perform the above rules until the board becomes stable, then return the stable board.
Example 1:
Input: board = [[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]] Output: [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]]
Example 2:
Input: board = [[1,3,5,5,2],[3,4,3,3,1],[3,2,4,5,2],[2,4,4,5,5],[1,4,4,1,1]] Output: [[1,3,0,0,0],[3,4,0,5,2],[3,2,0,3,1],[2,4,0,5,2],[1,4,3,1,1]]
Constraints:
m == board.lengthn == board[i].length3 <= m, n <= 501 <= board[i][j] <= 2000
Average Rating: 3.84 (50 votes)
Approach #1: Ad-Hoc
Intuition
We need to simply perform the algorithm as described. It consists of two major steps: a crush step, and a gravity step. We work through each step individually.
Algorithm
Crushing Step
When crushing, one difficulty is that we might accidentally crush candy that is part of another row. For example, if the board is:
123
145
111
and we crush the vertical row of 1s early, we may not see there was also a horizontal row.
To remedy this, we should flag candy that should be crushed first. We could use an auxillary toCrush boolean array, or we could mark it directly on the board by making the entry negative (ie. board[i][j] = -Math.abs(board[i][j]))
As for how to scan the board, we have two approaches. Let's call a line any row or column of the board.
For each line, we could use a sliding window (or itertools.groupby in Python) to find contiguous segments of the same character. If any of these segments have length 3 or more, we should flag them.
Alternatively, for each line, we could look at each width-3 slice of the line: if they are all the same, then we should flag those 3.
After, we can crush the candy by setting all flagged board cells to zero.
Gravity Step
For each column, we want all the candy to go to the bottom. One way is to iterate through and keep a stack of the (uncrushed) candy, popping and setting as we iterate through the column in reverse order.
Alternatively, we could use a sliding window approach, maintaining a read and write head. As the read head iterates through the column in reverse order, when the read head sees candy, the write head will write it down and move one place. Then, the write head will write zeroes to the remainder of the column.
We showcase the simplest approaches to these steps in the solutions below.
**Complexity Analysis**-
Time Complexity: O((R∗C)2), where R,C is the number of rows and columns in
board. We need O(R∗C) to scan the board, and we might crush only 3 candies repeatedly. -
Space Complexity: O(1) additional complexity, as we edit the board in place.
October 10, 2019 4:49 AM
This is just an implementation heavy problem; not an interesting one.
Last Edit: September 8, 2018 7:46 PM
Great solution!
It's O((R*C)^2) complexity because each function call scans the board three times so it's 3(R*C). If we only crush 3 candies each time, the function will be called (R*C)/3 times. Multiply those two terms together you get O((R*C)^2).
Last Edit: March 3, 2019 1:25 AM
Horrible code structure & no modular design at all...
(To be cleared, I like the explanation but wonder why it comes to a big mess in the actual implementation)
February 18, 2020 3:44 AM
Please dont use variable names like 'r' and 'R'. Makes code unreadable, will never pass code reviews (and also be frowned upon during interviews)
January 5, 2021 12:37 AM
I believe it should be a hard problem...
November 6, 2017 11:12 AM
@awice I have two questions: Firstly, although the function may call itself recursively, is it still true that we say it is a O(1) space? Secondly, would you please elaborate on time complexity O((R*C)^2). I do not get the square power completely.
Thanks
October 18, 2020 11:18 PM
This problem is so difficult, if I face it in interview, I will fail.
July 25, 2020 10:01 AM
don't know why people are complaining about the implementation, i think awice did it just fine this time. the code is pretty clear and concise.
November 12, 2020 1:21 AM
Is O(1) space complexity wrong? Since we may call the function recursively at most RC/3 times. Does is mean we need RC/3 stack space?
December 30, 2019 11:13 AM
Time complexity analysis explanation: Suppose there are R rows and C Columns. Then the number of vertical repetitions of length 3 we can crush are R/3 per column * C columns. The number of horizontal repetitions of length 3 we can crush are C/3 per row * R rows. Each crushing session takes O (R * C) time so we have O((R * C * R/3 * C) + (RCC/3R)) = O(R^2C^2/3 + R^2C^2/3) = O(R^2C^2).
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xxxxxxxxxxclass Solution {public: vector<vector<int>> candyCrush(vector<vector<int>>& board) { }};